Unique continuation property near a corner and its fluid-structure controllability consequences

DOI:10.1051/cocv:2008024 www.esaim-cocv.org

UNIQUE CONTINUATION PROPERTY NEAR A CORNER AND ITS FLUID-STRUCTURE

CONTROLLABILITY CONSEQUENCES

Axel Osses

and Jean-Pierre Puel

Abstract. We study a non standard unique continuation property for the biharmonic spectral prob- lem Δ²w=−λΔwin a 2D corner with homogeneous Dirichlet boundary conditions and a supplemen- tary third order boundary condition on one side of the corner. We prove that if the corner has an angle 0< θ0<2π,θ0=πandθ0= 3π/2, a unique continuation property holds. Approximate controllability of a 2-D linear fluid-structure problem follows from this property, with a control acting on the elastic side of a corner in a domain containing a Stokes fluid. The proof of the main result is based in a power series expansion of the eigenfunctions near the corner, the resolution of a coupled infinite set of finite dimensional linear systems, and a result of Kozlov, Kondratiev and Mazya, concerning the absence of strong zeros for the biharmonic operator [Math. USSR Izvestiya 34 (1990) 337–353]. We also show how the same methodology used here can be adapted to exclude domains with corners to have a local version of the Schiffer property for the Laplace operator.

Mathematics Subject Classification.35B60, 35B37.

Received September 19, 2005. Revised January 27, 2006 and September 27, 2007.

Published online March 28, 2008.

1. Introduction and main results

Let us consider a circular sector ofR² described in polar coordinates

G={(r, θ), 0< r < r₀, 0< θ < θ₀} (1.1) and centered at the origin. Let Ω be a Lipschitz bounded subset ofR² with a straight corner of angleθ₀at the origin, that is by definition: ifB_r0 is the open ball of radiusr₀ centered at the origin, then

Ω∩B_r0=G. (1.2)

Keywords and phrases. Continuation of solutions of PDE, fluid-structure control, domains with corners.

∗This work has been partially supported by ANR C-QUID 06BLAN0052, FONDECYT 1061263, STIC-AMSUD and ECOS C04E08 grants.

1Corresponding author: Departamento de Ingener´ıa Matem´atica and Centro de Modelamiento Matem´atico (UMI 2807 CNRS), FCFM Universidad de Chile, Casilla 170/3 - Correo 3, Santiago, Chile;[email protected]

2 Laboratoire de Math´ematiques de Versailles, UMR 8100, Universit´e de Versailles St-Quentin, 45 avenue des ´Etats-Unis, 78035 Versailles cedex, France;[email protected]

Article published by EDP Sciences c EDP Sciences, SMAI 2008

Letnbe the outward normal vector to Ω and let us take the notations

Γ₀={(r,0), 0< r < r₀}, Γ₁={(r, θ₀), 0< r < r₀}. (1.3) We are interested in the following local unique continuation property: prove (or disprove) that every weak solutionw∈H²(Ω) of the overdetermined spectral problem

Δ²w=−λΔw in Ω (1.4)

w=∂w

∂n = 0 on Γ₀∪Γ₁ (1.5)

∂Δw

∂n = 0 on Γ₀ (1.6)

necessarily vanishes in Ω. Our main result is the following:

Theorem 1.1. Let Ω⊂R² be a Lipschitz bounded subset with a straight corner of angle 0< θ₀<2π at the origin and assume that

θ₀=π, θ₀=3π

2 , (1.7)

then any weak solutionw∈H²(Ω) of the problem(1.4)–(1.6)vanishes in Ω.

We have become interested in this kind of problems because they are related to the approximate controlla- bility of some simplified fluid-structure models (see for instance [5] or [8]). Indeed, a non vanishing solution of (1.4)–(1.6) is the stream function of an eigenfunction of the Stokes operator with Dirichlet boundary condi- tions on Γ₀∪Γ₁ and havingconstant pressure on Γ₀. Theorem1.1says that such eigenfunction does not exist under condition (1.7). From this fact, a fluid-structure approximate controllability result can be obtained in the presence of a corner. In order to state the result, we consider the boundary of our domain Ω splitted into an elastic and a rigid part, that is

∂Ω = Γ_E∪Γ_R, Γ_E∩Γ_R=∅ (1.8)

and we introduce the Sobolev spaces

H ={z∈L²(Ω)² |divz= 0 in Ω, z·n= 0 on Γ_R} L²₀(Γ_E) ={ξ∈L²(Γ_E)|

ΓE

ξdσ= 0}.

With these definitions the result reads as follows:

Theorem 1.2. Let Ω be as in Theorem 1.1. We suppose that Γ_E and Γ_R are such that Γ₀ = Γ_E∩B_r0, Γ₁ = Γ_R∩B_r0. Given T > 0, u₀ ∈ L²(Ω), η₀ ∈ H₀²(Γ_E)∩L²₀(Γ_E) and η₁ ∈H₀¹(Γ_E)∩L²₀(Γ_E), let u, η be

solutions of the fluid-structure system

∂u

∂t −Δu+∇p= 0 inΩ×(0, T) (1.9)

divu= 0 inΩ×(0, T) (1.10)

u= 0 on Γ_R×(0, T) (1.11)

u=∂η

∂tn onΓ_E×(0, T) (1.12)

u(0) =u₀ inΩ (1.13)

∂²η

∂t² +Bη=−σ(u, p)n·n+h on Γ_E×(0, T) (1.14) η(t)∈H₀²(Γ_E)∩L²₀(Γ_E) a.e. in(0, T) (1.15) η(0) =η₀, ∂η

∂t(0) =η₁ onΓ_E, (1.16)

whereB is a selfadjoint differential operator which is uniformly elliptic inH₀²(Γ_E). Then, for anyu^T ∈L²(Ω), η^T₀ ∈H₀²(Γ_E)∩L²₀(Γ_E)andη^T₁ ∈H₀¹(Γ_E)∩L²₀(Γ_E)andε >0, there exists a function h∈L²(Γ_E)such that

u(T)−u^T _0,Ω+ η(T)−η^T_{0 2,Ω}+ η_t(T)−η^T_{1 1,Ω}< ε.

Remark 1.1. In [8] we have shown the well posedness of problem (1.9)–(1.16) by a transposition method.

The presence of the corner is important to obtain the results above. In fact, the local unique continuation property given in Theorem1.1is not true in a ball (see [5]).

The result of Theorem1.1 was already known for right corners (θ₀ =π/2), firstly for rectangular domains where a direct spectral solution of (1.4)–(1.6) can be obtained (see [7]) and secondly for general domains using the fact that forθ₀=π/2 the solution of (1.4)–(1.5) is analytic inG(see [8]). In this paper, we consider general corners except forθ₀= 3π/2 (andθ₀=πwhich does not correspond to a corner of course) by considering that w is not necessarily analytic but regular enough. We first prove in Section 2 that w isC^∞ in Gfor r₀ small enough using regularity results for the biharmonic operator near boundary corners. Then, in Sections3and4, by using a non trivial generalization of the technique used in [8], we show that all the derivatives ofwvanish at the origin,i.e.,whas a zero of infinite order. In Section5using a result of Kozlov, Kondratiev and Mazya [3], we deduce that this is only possible ifwvanishes.

The same method developed in the proof of Theorem1.1can be used to obtain some local unique continuation properties for the Laplace operator in domains with corners, which correspond in fact to a kind oflocal Schiffer’s conjecture(see for instance [1]).

More precisely, let Ω be an open subset ofR^N with unit exterior normalnand suppose that there exists an eigenvalueλand an eigenfunctionw∈H¹(Ω) satisfying

−Δw=λw in Ω (1.17)

∂w

∂n = 0 on∂Ω (1.18)

w= constant= 0 on Γ₀⊂∂Ω, (1.19)

for some Γ₀ strictlyincluded in ∂Ω. In this case we say that the domain Ω satisfies a local Schiffer property of Neumann type. The conjecture can be stated as:

Conjecture. The only simply connected domain satisfying a local Schiffer property of Neumann type is a ball.

The analogous local Schiffer conjecture of Dirichlet type is obtained by interchanging the role of Neumann and Dirichlet boundary conditions in (1.18)–(1.19).

If Γ₀=∂Ω or∂Ω is analytic, then condition (1.19) holds on the whole boundary and we are in the case of classical Schiffer conjectures. In this context, it is well known that the classical Schiffer conjecture of Neumann type is equivalent to say that the ball is the only simply connected domain having the Pompeiu property [9].

The general case Γ₀ = ∂Ω is different and, to our knowledge, a relationship between the local Schiffer property of Neumann type and some kind of restricted Pompeiu property is not known. But, there is another characterization that admits a local version. Indeed, in the case of simple eigenvalues, we can rewrite the local Schiffer property of Neumann type as an extremal Neumann eigenvalue problem under volume constraint as in [1]. That is, ifλ_n is a simple Neumann eigenvalue of (1.17)–(1.18) associated to a normalized eigenvectorw_n in L²(Ω), then

dλ(Ω;v) =

Γ0

(v·n)

λ_nw²_n− ∂w_n

∂n ²

dσ

gives its derivative with respect to the domain for a regular deformation fieldvwhichvanishesin∂Ω\Γ₀. The characterization is that (λ_n, w_n) is a solution of (1.17)–(1.19) iff there exists a constantcsuch that

dλ_n(Ω;v) =c

Γ0

(v·n) dσ, for all deformationv withv|_∂Ω\Γ0 = 0.

The proof of this equivalence is exactly the same as in [1], Proposition 2.2, after replacing∂Ω by Γ₀.

Here we prove a partial answer to the previous local Schiffer conjecture of Neumann type in the caseN = 2:

Theorem 1.3. Let Ω be as in Theorem 1.1. If there exists a pair(λ, w)∈R×H¹(Ω) satisfying (1.17),(1.18) and(1.19), thenwnecessarily vanishes in Ω.

The proof of this result is given in Section 6.

In the case Γ₀=∂Ω or∂Ω analytic, Theorem1.3 becomes a particular case of a more general result due to Williams [10], saying that a Lipschitz domain with a boundary which is not analytic everywhere does not have the Pompeiu property.

2. Local regularity at the origin

We recall the followingH^m+3-regularity result for the biharmonic operator near a corner (see Grisvard [2], Th. 7.2.2.3 and Rem. 7.2.2.4):

Theorem 2.1 (Grisvard [2]). Given a corner G_∞={(r, θ), r >0, 0< θ < θ₀},f ∈H^m(G_∞),m≥0, ifwis a bounded support solution of

Δ²w=f in G_∞, w∈H₀²(G_∞) (2.1)

then

w=w_r+w_sη (2.2)

wherew_r∈H^m+3(G_∞),η is aC^∞(G_∞)cut-off function equal to1 near the origin and independent of mand

w_s=

−(m+1)≤pk<0

r^1+ipka_ku_pk(θ) +

−(m+1)≤q<0

r^1+iq

bu_q(θ) +c

v_q(θ) +i(lnr)u_q(θ) , (2.3)

wherea_k,b,c are complex constants and p_k,q are respectively the simple and double rootsτ of

sinh²(τθ₀) =τ²sin²(θ₀) (2.4)

with imaginary part in [−(m+ 1),0) and excluding the root −i if tanθ₀=θ₀. The functionsu_pk,u_q, v_q can be chosen uniquely prescribed as solutions of the linear fourth order ordinary differential equations

u^(iv)_τ + 2(1−τ²)u_τ+ (1 +τ²)²u_τ= 0 in(0, θ₀) (2.5) u_τ(0) =u_τ(θ₀) =u_τ(0) =u_τ(θ₀) = 0, u_{τ L}2(0,θ0)= 1, (2.6) corresponding both toτ =p_k andτ=q and

v^(iv)_τ + 2(1−τ²)v_τ+ (1 +τ²)²v_τ= 4τu_τ−4τ(1 +τ²)u_τ in(0, θ₀) (2.7) v_τ(0) =v_τ(θ₀) =v_τ(0) =v_τ(θ₀) = 0, (u_τ, v_τ)_L2(0,θ0)= 0, (2.8) corresponding only to τ=q.

Remark 2.1. A classicalH^m+4 regularity result holds for a more restrictive set of anglesθ₀. More precisely, if in the previous theorem, all p_k and q_l have imaginary part different from −(m+ 2) then w_r ∈H^m+4(G_∞) and an analogous decomposition as (2.3) holds with sums over imaginary part in the range [−(m+ 2),0). It is important to use here theH^m+3 regularity result to avoid unnecessary restrictions onθ₀.

First we will prove the following regularity result:

Theorem 2.2. Let Ω⊂R² be a Lipschitz bounded domain with a straight corner of angle 0< θ₀ <2πat the origin, then any weak solution w∈H²(Ω)of the problem (1.4)–(1.6)isC^∞ at the origin.

In order to prove Theorem 2.2 we will need some extra auxiliary lemmas. The first one is obtained by localization in an standard way as in [2].

Lemma 2.1. If w∈H²(Ω)is solution of (1.4)–(1.5)thenw∈C^∞(G_∞\B_δ),∀δ >0.

The second one concerns some particular properties of the singular part of the solution:

Lemma 2.2. (i)The functions r^1+ipk,r^1+iq,r^1+iqlnr appearing in(2.3)are linearly independent.

(ii)The singular component w_s given by the expansion (2.3)is a biharmonic function.

Proof. (i) It is clear, by definition, that the roots p_k, q of the characteristic equations are all different so the above functions are clearly independent. (ii) The fact that Δ²w_s = 0 is implicitly given by construction in [2], Section 7.2. Nevertheless, we give here an explicit proof in order to be self-contained. Letr = e^t and z_s(t, θ) = e^−tw_s(e^t, θ). We have

Δ²w_s= e^−3tP(D_t, D_θ)z_s

where D_t,D_θ stand for the derivatives with respect to t andθ respectively and the differential operator P is given by

P(D_t, D_θ) = (D_t²−1)²+ 2(D²_t+ 1)D²_θ+D⁴_θ. From (2.3), it follows that z_sis a linear combination of the functions

e^iτtu_τ and e^iτt(v_τ+itu_τ),

where τ is a simple or double root of (2.4) with imaginary part in [−(m+ 1),0), excluding the root −i if tanθ₀=θ₀. The functionsu_τ,v_τ are the unique solutions of (2.5), (2.6) and (2.7), (2.8) respectively. If

p(τ, D_θ) =P(iτ, D_θ) = (τ²+ 1)²+ 2(1−τ²)D²_θ+D_θ⁴

it is easy to verify that

P(D_t, D_θ)e^iτtu_τ= e^iτtp(τ, D_θ)u_τ= 0 P(D_t, D_θ)e^iτt(v_τ+itu_τ) = e^iτt

it p(τ, D_θ)u_τ+p(τ, D_θ)v_τ+_∂τ^∂ p(τ, D_θ)u_τ

= 0,

where we have used (2.5) and (2.7). We conclude that Δ²w_s= 0.

Lemma 2.3. Letτbe a simple or double root of(2.4)with imaginary part in[−(m+1),0), excluding the root−i if tanθ₀=θ₀. Letu_τ be the corresponding solution of(2.5)with boundary conditions (2.6). If u_τ(0) = 0 then u_τ is zero.

Proof. The casesτ= 0,τ = +i are not allowed since the imaginary part ofτ is strictly negative. So, the only cases to consider are: (i) τ=±i,τ= 0 if tanθ₀=θ₀or (ii) τ=−iif tanθ₀=θ₀.

In case (i), the characteristic values of (2.5) areτ+i, τ−i, −τ−i, −τ+i, so the solution is of the form u_τ =Aexp((τ+i)θ) +Bexp((τ−i)θ) +Cexp((−τ−i)θ) +Dexp((−τ+i)θ). In case (ii), the characteristic values of (2.5) are 0 (double) and 2i,−2i, so the solution is of the formu_τ=A+Bθ+Cexp(2iθ) +Dexp(−2iθ).

In both cases, and for the boundary conditions (2.6), the corresponding linear homogeneous system associated to the coefficientsA, B, C andD has vanishing determinant, since of courseτ verifies (2.4) to have subspaces of nontrivial solutions. Indeed, in case (i) the determinant is Δ₁ = 4(sinh(τθ₀)−τ²sin²(θ₀)) which vanishes due to (2.4) and in case (ii), it is Δ₂= 16isin(θ₀)(θ₀cosθ₀−sinθ₀) which vanishes since tanθ₀=θ₀.

Now, if we add the overdetermined condition u_τ(0) = 0 and we replace the first row of the corresponding matrices associated to the previous linear homogeneous system with this condition, we obtain the non vanishing determinantsΔ₁= 16τ(1 +τ²)(sinh(2τθ₀) +τsin(2θ₀)) or using (2.4)Δ₁= 32τ(1 +τ²) sinh(τθ₀)(cosh(τθ₀)± cosθ₀) in the first case andΔ₂=−4 sin(θ₀) in the second case.

In both cases we obtain the trivial solution,i.e. A=B =C=D= 0.

The other two lemmas are related to regularity properties.

Lemma 2.4. Let sbe a non-negative real number, α∈C,β ∈ {0,1} andϕ∈C^∞([0, θ₀]), and let η be a C^∞ cut-off function equal to one near the origin. Let D=G_∞ (in this case we setN = 2) orD =∂G_∞∩ {θ= 0}

(in this case we set N = 1). Then

r^α(lnr)^βϕ(θ)η|_D∈H^s(D)⇔

α > s−N 2

or (α∈Nandβ= 0) or ϕ|_D= 0

.

Proof. LetV be a neighborhood of the origin whereη= 1 in such a way to consider theH^sregularity inD∩V in which case we can drop off the function η. Ifs=k is a non-negative integer, thenD^γ(r^α(lnr)^βϕ(θ)|D∩V) for|γ| ≤k, is a linear combination of functions of the formr^α−m(lnr)^β andr^α−m for 0≤m≤kwhen N= 1 or r^α−m(lnr)^βψ(θ) andr^α−mφ(θ), where 0 ≤m ≤k andψ and φ areC^∞([0, θ₀]) whenN = 2. This occurs except for the case α∈ N and β = 0 when r^α is a polynomial in r or in the trivial case where ϕ|_D∩V = 0.

Assuming we are not in these cases, the function is inH^s(D∩V) if and only if 2(α−k) +N >0, which gives the characterization of the lemma.

For non-integersthis follows as in [2], Theorem 1.4.5.3, by an embedding method due to Babuˇska. Take a non-negative integer k≤ s and p≥2 such that k−^N_p = s− ^N₂. Since α≤k− ^N_p then the functions are not inW_p^k(D∩V), except ifα∈Nand β = 0 orϕ|D∩V = 0, and the Sobolev imbedding ofH^s(D∩V) into

W_p^k(D∩V) gives the result.

Lemma 2.5 (see [4]). LetΩ be a Lipschitz domain. Ifu∈H¹(Ω) andΔu∈L²(Ω) then ^∂u_∂n ∈H^−1/2(∂Ω).

Let us now proceed with the proof of Theorem2.2. Let Ω,r₀,G, Γ₀, Γ₁be given by (1.1), (1.2) and (1.3).

First of all, we will localize problem (1.4)–(1.5)–(1.6) near the origin. Letξ =ξ(r) be a C^∞ cut-off function

such that ξ = 1 for 0 < r < ¹₃r₀ and ξ = 0 for r > ²₃r₀. By defining w = ξ w if (r, θ) ∈ G and w = 0 if (r, θ)∈G_∞\G, it is easy to see that it satisfies

w∈H₀²(G_∞) and Δ²w=−λΔw+h, where

h=λΔξw+ 2λ∇ξ· ∇w+ 2∇ξ· ∇Δw+ 2ΔξΔw+ 2Δ(∇ξ· ∇w) + 2∇Δξ· ∇w+ Δ²ξ w.

Notice that h∈C^∞(G_∞) and vanishes forr < r₀/3 since∇ξvanishes in this domain andwis C^∞ out of this region (Lem. 2.1). Therefore, we can apply Theorem2.1withm= 0 to the functionwsince it is a compactly supportedH₀²(G_∞) solution of

Δ²w=−λΔw+h=f ∈L²(G_∞).

Therefore, we have the decompositionw =w_r+w_sη, wherew_r ∈ H³(G_∞), η is a C^∞(G_∞) cut-off function equal to 1 near the origin (say, forr < r₁< r₀/3) and

w_s=

−1≤pk<0

r^1+ipka_ku_pk(θ) +

−1≤q<0

r^1+iq(b_lu_q(θ) +c_l(v_q(θ) +i(lnr)u_q(θ))). (2.9)

The idea is to prove that all the complex coefficients in the previous expansion vanish, except possibly for those a_k,b associated to rootsp_k,qsuch that 1 +ip_kor 1 +iqis a non-negative integer. In fact, we will exclude the H³(G_∞) terms associated to these roots fromw_sby supposing they are already included in the regular partw_r. Notice that Δ²w_s= 0 (see Lem.2.2) and therefore Δ²(w_sη) = 0 forr < r₁. Then Δ²w_r= Δ²w−Δ²(w_sη) = f ∈L²(G_∞∩B_r1). Also Δw_r∈H¹(G_∞∩B_r1) since w_r∈H³(G_∞∩B_r1), so by Lemma2.5 we have

∂Δw_r

∂n ∈H^−1/2((Γ₀∪Γ₁)∩B_r1).

The space H^−1/2((Γ₀∪Γ₁)∩B_r1) can not be restricted to H^−1/2(Γ₀∩B_r1). We are only interested in the regularity of ^∂Δw_∂n^r near the origin. In order to avoid the other part of the boundary of Γ₀∩B_r1 we use another C^∞ cut-off functionη₁ equal to one forr < r₁/3 and vanishing forr >2r₁/3. We claim that

η₁∂Δw_r

∂n ∈

H₀₀^1/2(Γ₀∩B_r1) . (2.10)

Indeed, the functions of H₀₀^1/2(Γ₀∩B_r1) can be extended by zero continuously toH^1/2((Γ₀∪Γ₁)∩B_r1). For the definition of the trace spaceH₀₀^1/2 see [4]. One characterization of this space is

f ∈H₀₀^1/2(Γ₀∩B_r1)⇔f ∈H^1/2(Γ₀∩B_r1) andr^−1/2(r−r₁)^−1/2f ∈L²(Γ₀∩B_r1).

Using functions with behavior like _ln¹_r nearr= 0, which belong toH₀₀^1/2, we can show that r^α∈

H₀₀^1/2(Γ₀∩B_r1) ⇔ α≤ −1. (2.11)

Sincew=wforr < r₁, the overdetermined boundary condition (1.6) also holds forwifr < r₁ and therefore η₁∂Δw_s

∂n =η₁∂Δw

∂n −η₁∂Δw_r

∂n =−η₁∂Δw_r

∂n ∈

H₀₀^1/2(Γ₀∩B_r1) .

Using the formula for the Laplacian in polar coordinates we compute

∂Δ(f(r)g(θ))

∂n = 1

r

∂

∂θ 1

rf+f

g+ 1 r²fg

= 1

r 1

rf+f

g+ 1 r³fg and ifg(0) = 0 we obtain

∂Δ(f(r)g(θ))

∂n

θ=0

= 1

r³fg(0).

Using this in (2.9) and recalling the boundary conditions (2.6) and (2.8) we obtain on Γ₀ forr < r₁/3

∂Δw_s

∂n (r) =

−1≤pk<0

r^−2+ipka_ku_pk(0) +

−1≤q<0

r^−2+iq

b_lu_q(0) +c_l

v_q(0) +i(lnr)u_q(0) .

From the previous expansion and the characterization (2.11) we verify that η₁∂Δw_s

∂n ∈

H₀₀^1/2(Γ₀∩B_r1) , since the real partαof the powers of rsatisfies

α≤ −2 + 1 =−1.

From (2.10) the only possibility is

η₁∂Δw_s

∂n

Γ0

= 0.

We recall that the functions

r^−2+ipk, r^−2+iq, r^−2+iqlnr are linearly independent, therefore we get

a_ku_p

k(0) = 0 ∀p_k c_lu_q(0) = 0 ∀q b_lu_q(0) +c_lv_q(0) = 0 ∀q. From the first equation, ifa_k = 0 for some k thenu_p

k(0) = 0. This is an overdetermined condition for (2.5)–

(2.6), which would implyu_pk= 0. But this is a contradiction due to the normalized condition appearing in (2.6).

Then a_k = 0 for allk. Now, from the second equation, ifc_l = 0 for somel then u_q(0) = 0. This is also an overdetermined condition for (2.5)–(2.6) and would implyu_q

= 0, which gives a contradiction. Thereforec_l= 0 for alll. Finally, from the third equation, ifb_l= 0 thenu_q(0) = 0 and this gives again a contradiction. Notice that the functionv_q, which is solution of a non homogeneous equation (2.7)–(2.8), does not intervene in this analysis.

Consequently

w_s= 0 and therefore

w=w_r∈H³(G∩B_r1).

The next steps are easier. Let us suppose now thatw∈H^m+2(G_∞) for somem≥1. Then we apply Theorem2.1 to obtain that w = w_r+w_sη with w_r ∈H^m+3(G_∞) andw_s of the form (2.3). With the same argument as before, it is easy to verify that

∂Δw_s

∂n = ∂Δw

∂n −∂Δw_r

∂n =−∂Δw_r

∂n ∈H^m−1/2(Γ₀∩B_r1).

But

r^α(lnr)^βη ∈H^m−1/2(Γ₀∩B_r1) if α≤m−1, β= 0 or 1,

by Lemma 2.4, except if α is a non-negative integer and β = 0 in which case the functions are H^m+3 and they are already included in w_r. In the present case, the real part of the powers of r in ^∂Δw_∂n^s are given by α=(1 +iτ−3) =−2− τ,τ =p_k or τ=q and they satisfy

1<α≤m−1 sinceτ∈[−(m+ 1),0). Therefore

∂Δw_s

∂n = 0 on Γ₀∩B_r1

and the same argument of linear independence as before shows that this is possible only ifw_s vanishes. Then

w=w_r∈H^m+3(G_∞).

So we obtain thatwis inH^m+3(G_∞) for allm≥0 and this gives theC^∞regularity at the origin. This finishes the proof of Theorem2.2.

3. Power series expansion

We choose Cartesian axes (x, y) centered inr= 0,θ= 0 where the edge Γ₀ofGcoincides with the horizontal axis. Sincew∈C^∞(G∩B_ρ) for someρ >0, for (x, y) in a neighborhood of (0,0), we can write for eachk≥0

w(x, y) =

i,j≥0 i+j≤k+4

a_ijxⁱy^j+o(x^k+4+y^k+4). (3.1)

First derivatives give:

∂w

∂x =

i≥1,j≥0 i+j≤k+4

i a_ijxⁱ⁻¹y^j+o(x^k+3+y^k+3)

∂w

∂y =

i≥0,j≥1 i+j≤k+4

j a_ijxⁱy^j−1+o(x^k+3+y^k+3)

Δw =

i≥2,j≥0 i+j≤k+4

i(i−1)a_ijxⁱ⁻²y^j+

i≥0,j≥2 i+j≤k+4

j(j−1)a_ijxⁱy^j−2+o(x^k+2+y^k+2),

from where

∂w

∂x =

i,j≥0 i+j≤k+3

(i+ 1)a_i+1,jxⁱy^j+o(x^k+3+y^k+3) (3.2)

∂w

∂y =

i,j≥1 i+j≤k+3

(j+ 1)a_i,j+1xⁱy^j+o(x^k+3+y^k+3) (3.3)

Δw=

i,j≥0 i+j≤k+2

(i+ 2)(i+ 1)a_{i+2, j}+ (j+ 2)(j+ 1)a_i,j+2 xⁱy^j+o(x^k+2+y^k+2) (3.4)

Δ²w=

i,j≥0 i+j≤k

(i+ 4)(i+ 3)(i+ 2)(i+ 1)a_i+4,j+ 2(i+ 2)(i+ 1)(j+ 2)(j+ 1)a_i+2,j+2

+ (j+ 4)(j+ 3)(j+ 2)(j+ 1)a_i,j+4 xⁱy^j+o(x^k+y^k). (3.5) Note that the expansions (3.1) to (3.5) are valid for allk≥0. We can then rewrite equation (1.4) by identifying coefficients in (3.4)–(3.5) as

(i+ 4)!j!a_i+4,j+ 2(i+ 2)!(j+ 2)!a_i+2,j+2+i!(j+ 4)!a_i,j+4=

−λ

(i+ 2)!j!a_i+2,j+i!(j+ 2)!a_i,j+2 , ∀i, j≥0. (3.6) Boundary conditions (1.5)–(1.6) on Γ₀ are also equivalent to

a_i,0= 0, a_i,1= 0, a_i,3= 0, ∀i≥0. (3.7)

Now we introduce a parameterα= tanθ₀ such that

(x, y)∈Γ₁ implies α= tanθ₀. (3.8)

Note that, from hypothesis (1.7) and since the caseθ₀=π/2 was specially treated in [8], we can assume that α= 0, α=±∞.

The first boundary condition (1.5) on Γ₁ becomes

w(x, αx) =

i,j≥0 i+j≤k+4

a_ijα^jx^i+j+o(x^k+4)

=

k+4

=0

i,j≥0 i+j=

a_ijα^j

x+o(x^k+4)

= 0

and then

i,j≥0 i+j=k

a_ijα^j= 0 ∀k≥0. (3.9)

Since on Γ₁ the normal isn= (α,−1)/√

1 +α² and the tangentτ = (1, α)/√

1 +α² then

∂w

∂n = 1

√1 +α²

α∂w

∂x −∂w

∂y

(3.10)

∂w

∂τ = 1

√1 +α² ∂w

∂x +α∂w

∂y

= 0, (3.11)

and the other boundary condition (1.5) on Γ₁can be replaced by

∂w

∂x(x, αx) =

i,j≥0 i+j≤k+3

(i+ 1)a_i+1,jα^jx^i+j+o(x^k+3)

=

k+3

=0

i,j≥0 i+j=

(i+ 1)a_i+1,jα^j

x+o(x^k+3)

= 0

that is

i,j≥0 i+j=k

(i+ 1)a_i+1,jα^j = 0 ∀k≥0. (3.12)

Now we have replaced equations (1.4)–(1.6) by (3.6)–(3.12).

The idea is first to show that

a_i,2k+1= 0 ∀k≥0, ∀i≥0. (3.13)

This is easily shown by induction since from (3.7) a_i,1 = a_i,3 = 0, ∀i ≥0 and if we suppose that a_i,2k−3 = a_i,2k−1= 0 equation (3.6) forj = 2k−3 gives

i!(2k+ 1)!a_i,2k+1 = 0 from which we deduce (3.13).

The hardest technical part is to show that

a_i,2k= 0 ∀k≥0, ∀i≥0, (3.14)

and this is given in the next section.

4. An infinite set of finite dimensional systems

Let us show (3.14) by induction. From (3.7) we know that a_i,0 = 0,∀i≥0 and from (3.13) we know that a_i,2k+1= 0,∀k≥0. Using (3.6) with both i+j=k= 2,3 we obtain

a_0,2α²+a_1,1α+a_2,0= 0 ⇒ a_0,2= 0 (4.1) a_0,3α³+a_1,2α²+a_2,1α+a_3,0= 0 ⇒ a_1,2= 0. (4.2) Now, suppose that

a_2k+2,0=a_2k,2=. . .=a_2,2k =a_0,2k+2= 0 (index sum 2k+ 2) (4.3) a_2k+1,0=a_2k−1,2=. . .=a_3,2k−2=a_1,2k= 0 (index sum 2k+ 1) (4.4)

let us prove that

a_2k+4,0=a_2k+2,2=. . .=a_2,2k+2 =a_0,2k+4= 0 (index sum 2k+ 4) (4.5) a_2k+3,0=a_2k+1,2=. . .=a_3,2k=a_1,2k+2= 0 (index sum 2k+ 3). (4.6) Taking i = 2k, j = 0, i = 2k−2, j = 2,. . ., i = 2, j = 2k−2,i = 0, j = 2k in order to have a zero right hand side in (3.6) (index sum 2k+ 2), if we also take into account (3.7) we obtain the following system ofk+ 1 equations which do not depend onλexplicitly:

2(2k+ 2)!2!a_2k+2,2+ (2k)!4!a_2k,4 = 0 (4.7) (2k+ 2)!2!a_2k+2,2+ 2(2k)!4!a_2k,4+ (2k−2)!6!a_2k−2,6 = 0 (4.8)

...

4!2k!a_4,2k+ 2·2!(2k+ 2)!a_2,2k+2+ 0!(2k+ 4)!a_0,2k+4 = 0. (4.9) This system has k+ 2 unknowns,i.e. the vector:

A_k+2= (a_2k+2,2, a_2k,4, . . . , a_2,2k+2, a_0,2k+4).

In fact the system is overdetermined by the additional two conditions (3.9) and (3.12). More precisely, if we takei+j= 2k+ 4 in (3.9) andi+j= 2k+ 3 in (3.12) we respectively obtain

α²a_2k+2,2+α⁴a_2k,4+. . .+α^2k+2a_2,2k+2+α^2k+4a_0,2k+4 = 0 (4.10) (2k+ 2)α²a_2k+2,2+ (2k)α⁴a_2k,4+. . .+ 2α^2k+2a_2,2k+2 = 0. (4.11) If we introduce the variable

β = 1

α − ∞< β <+∞

it is clear thatA_k+2 satisfies simultaneously the systems

M_k+2A_k+2= 0, N_k+2A_k+2= 0, (4.12)

whereM_k+2andN_k+2are both squarek+ 2×k+ 2 matrices,M_k+2 including the coefficients of (4.7) and (4.10) andN_k+2 including the coefficients of (4.7) and (4.11),i.e.

M_k+2=

⎛

⎜⎜

⎜⎜

⎜⎝

2(2k+ 2)!2! (2k)!4! 0 . . . . . . 0 0

(2k+ 2)!2! 2(2k)!4! (2k−2)! 0 . . . 0 0

... . .. ... . .. ... . .. ...

0 . . . . . . 0 4!(2k)! 2·2!(2k+ 2)! (2k+ 4)!

β^2k+2 β^2k . . . . . . β⁴ β² 1

⎞

⎟⎟

⎟⎟

⎟⎠

and

N_k+2 =

⎛

⎜⎜

⎜⎜

⎜⎝

2(2k+ 2)!2! (2k)!4! 0 . . . . . . 0 0

(2k+ 2)!2! 2(2k)!4! (2k−2)! 0 . . . 0 0

... . .. ..

. . .. ..

. . .. ..

. 0 . . . . . . 0 4!(2k)! 2·2!(2k+ 2)! (2k+ 4)!

(2k+ 2)β^2k+2 (2k)β^2k . . . . . . 4β⁴ 2β² 1

⎞

⎟⎟

⎟⎟

⎟⎠ .

We have to prove that

D_k+2(β) = detM_k+2

(2k+ 2)!2!(2k)!4!. . .2!(2k+ 2)!0!(2k+ 4)!, (4.13)

E_k+2(β) = detN_k+2

(2k+ 2)!2!(2k)!4!. . .2!(2k+ 2)!0!(2k+ 4)! (4.14) can not vanish simultaneously. It is clear that

D_k+2 =

2 1

1 2 1

1 2 1

. .. ... ...

1 2 1

1 2 1

β^2k+2

(2k+2)!2! β^2k

(2k)!4! . . . . _2!(2k+2)!^β2 _0!(2k+4)!¹ and

E_k+2=

2 1

1 2 1

1 2 1

. .. ... ...

1 2 1

1 2 1

(2k+2)β^2k+2

(2k+2)!2! (2k)β^2k

(2k)!4! . . . . _2!(2k+2)!^2β2 _0!(2k+4)!⁰ .

From the formula for the derivative of a determinant it is clear that E_k+2 =β ∂

∂βD_k+2, so assuming that for some parameterβ₀ we have

D_k+2(β₀) = 0, E_k+2(β₀) = 0 this implies that

D_k+2(β₀) = 0, ∂

∂βD_k+2(β₀) = 0, that is to say,β₀ is a double root ofD_k+2.

In factD_k+2can be calculated explicitly by developing the determinant (4) with respect to the last column, giving the following recursive formula

D_k+2= 1

0!(2k+ 4)!Δ_k+1−β²D_k+1, where

Δ_k+1=

2 1

1 2 1

. .. ... ...

1 2 1

1 2

=k+ 2.

From the above recurrence, it is possible to deduce that D_k+2 =

k+1

j=0

(−1)^jβ^2j(k+ 2−j) (2j)!(2(k−j) + 4)!

= 1

2(2k+ 3)!Re(1 +iβ)^2k+3. (4.15)

Notice thatD_k+2(β) is a polynomial of degree 2k+ 2 inβ. If ω= arg(1 +iβ), ω ∈

−π 2,π

2 ,

then the number of different roots ofD_k+2corresponds to the number of different arguments in

−^π₂,^π₂

solutions of the equation

(2k+ 3)ω=π

2 +π, ∈Z.

These different solutions are

w= π

2(2k+ 3)(1 + 2), −(k+ 1)≤≤k

, (4.16)

that is, exactly 2k+ 2 different values in

−^π₂,^π₂

. Therefore, all the roots ofD_k+2 are distinct and they are not double roots.

From the above analysis, the only solution of systems (4.12) is the trivial one:

A_k+2= (a_2k+2,2, a_2k,4, . . . , a_2,2k+2, a_0,2k+4) = (0,0, . . . ,0,0).

With an analogous technique, it is possible to show that

B_k+2= (a_2k+1,2, a_2k−1,4, . . . , a_3,2k+2, a_1,2k+4) = (0,0, . . . ,0,0).

This shows that all the coefficients of the Taylor expansion ofwsolution of (1.4)–(1.6) near the origin are zero, so that the origin is a zero of infinite order ofw.

5. A zero of infinite order

We use the following result of Kozlov, Kondratiev and Mazya about the zeros of infinite order for the biharmonic operator. First we introduce the spaceV_n^k(G) as the space of functions defined inGfor which

r^{(−k+|α|+n)}D^αw∈L²(G), |α| ≤k.

Theorem 5.1 ([3]). Supposeθ₀=πandθ₀= 2πandw∈V₀⁴(G)is solution of the differential inequality

|Δ²w| ≤ C r²

|Δw|+1 r|w|

forr < r₀, 0< θ < θ₀ (5.1) w(r,0) =w(r, θ₀) =^∂w_∂θ(r,0) = ^∂w_∂θ(r, θ₀) = 0 forr < r₀ (5.2) and suppose also that

w∈V_n⁴(G), ∀n≤ −1 (5.3)

thenw= 0in G∩B_r0.

It is clear that if (λ, w) is solution of (1.4)–(1.5) andλis fixed, thenwsatisfies (5.1)–(5.2) forr₀small enough.

From the previous sections we know that w∈ C^∞(G∩B_ρ) for some ρ >0 and that all the derivatives of w vanish at the origin, which implies that (5.3) holds. We then necessarily have

w= 0 in G∩B_ρ,

for some ρ > 0 sufficiently small. Finally, by standard unique continuation we deduce thatw vanishes in G.

This ends the proof of Theorem1.1.

6. Non standard unique continuation for the Laplace operator in a corner

This section gives the proof of Theorem 1.3 which essentially says that domains with corners do not have the local Schiffer property of Neumann type defined in the Introduction. For this aim, we prove a non standard local unique continuation for solutions of (1.17)–(1.19) in a cornerGof the form (1.1) forming an angleθ₀ and verifying (1.7).

The proof follows the same steps of the proof of Theorem1.1, but it is simpler. First, from classical results of elliptic regularity near corners (see [2,6]), it is possible to prove that the solutionwof (1.17)–(1.19) is in fact C^∞ at the origin. This is done as in Section2 after proving that the eventually singular part of the Neumann eigenfunctions vanishes thanks to the overdetermined condition (1.19). We have to use aH^m+2−εm, regularity result withε_m>0 chosen in order to avoid corner’s angle restrictions and after a bootstrap argument to obtain theC^∞regularity at the origin. Then we expandwin finite series near the origin of the corner for eachk≥0

w(x, y) =

i,j≥0 i+j≤k+2

a_ijxⁱy^j+o(x^k+2+y^k+2). (6.1)

As done in Section3, and introducing the slopeα= tanθ₀ of Γ₁= (∂Ω\Γ₀)∩B_r0, (B_r0 from Th.1.3) equations (1.17)–(1.19) can be rewritten as

(i+ 2)(i+ 1)a_i+2,j+ (j+ 2)(j+ 1)a_i,j+2=−λ a_ij ∀i, j≥0 (6.2)

i,j≥0 i+j=k

α^j

α(i+ 1)a_i+1,j−(j+ 1)a_i,j+1 = 0 ∀k≥0, (6.3)

a_i,1= 0 ∀i≥0 (6.4)

a_0,0=c, a_i,0= 0 ∀i≥1 (6.5)

where c = 0 is the constant appearing in condition (1.19). Condition (6.2) comes from formula (3.4) and condition (6.3) is deduced from the expression for the normal derivative on Γ₁ given by (3.10) and partial derivatives (3.2)–(3.3). It is easy to see that from the above conditions we necessarily have c = 0. Indeed, from (6.2), (6.4) and (6.5) it is easy to see that all coefficientsa_ij vanishes, except eventually for

a_0,0=c and a_0,2k ∀k≥1.

But from (6.2) withi=j= 0 and from (6.2) withk= 2 we respectively have 2a_0,2=−λ c and 2α a_0,2= 0.

Sinceλ= 0 andα= 0, we would havec= 0 and this would be possible only ifw= 0.